Why does the first main function need "%s" in order to print out the text and second one does not?

Question

#include <stdio.h>

int main() {
int score = 89;

if (score >= 90)
printf("%s", "Top 10% \n");
else if (score >= 80)
printf("%s", "Top 20% \n");
else if (score > 75)
printf("%s", "You passed.\n");
else
printf("%s", "You did not pass.\n");
}

#include <stdio.h>

int main() {
int num = 41;
num += 1;
if (num == 42) {
printf("You won!");
}
}

Answer 1

Actually, there is a difference.

The first parameter is parsed by printf, where it looks for format specifiers.

What if that first string contains a substring like %s, %d or any other format? Bad things, that's what.
When it hits something like %s, it expects that there is another parameter passed to it, so it will read the second parameter, even if you didn't pass anything.

Consider this string:
char s[] = "Hello %s";

If we print it with:
printf( "%s", s );
we get the expected: "Hello %s"

But what if we did this instead?
printf( s );
In my case it prints "Hello -m7"
Sometimes: "Hello "
or "Hello (NULL)"

This is undefined behaviour.

So, never pass your string as the first parameter, unless it's purpose is to specify the format for the other parameters.
Never, ever pass a string as the first parameter that can be modified by the user.
( The user may enter a string that can be used to read out data from the stack which can be used in a malicious way )

Answer 2

You can use both print function in main. There is no any exceptional case. These are two ways to print String value in C

You can see here

https://www.sololearn.com/learn/C/2914/